(简单dp) poj 3356 AGTC

最新推荐文章于 2019-08-11 00:53:32 发布

Hashmat

最新推荐文章于 2019-08-11 00:53:32 发布

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分类专栏：动态规划文章标签： string integer input output c user

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本文介绍了一道经典的字符串匹配问题，通过动态规划求解两个字符串之间的最小编辑距离，即从一个字符串转换到另一个字符串所需的最少操作数（包括插入、删除和替换）。文章提供了完整的C++代码实现，并解释了算法的基本思路。

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AGTC

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5058		Accepted: 1953

Description

Let x and y be two strings over some finite alphabet A. We would like to transform x into y allowing only operations given below:

Deletion: a letter in x is missing in y at a corresponding position.
Insertion: a letter in y is missing in x at a corresponding position.
Change: letters at corresponding positions are distinct

Certainly, we would like to minimize the number of all possible operations.

Illustration
A G T A A G T * A G G C

| | |       |   |   | |

A G T * C * T G A C G C
Deletion: * in the bottom line
Insertion: * in the top line
Change: when the letters at the top and bottom are distinct

This tells us that to transform x = AGTCTGACGC into y = AGTAAGTAGGC we would be required to perform 5 operations (2 changes, 2 deletions and 1 insertion). If we want to minimize the number operations, we should do it like

A  G  T  A  A  G  T  A  G  G  C

|  |  |        |     |     |  |

A  G  T  C  T  G  *  A  C  G  C

and 4 moves would be required (3 changes and 1 deletion).

In this problem we would always consider strings x and y to be fixed, such that the number of letters in x is m and the number of letters in y is n where n ≥ m.

Assign 1 as the cost of an operation performed. Otherwise, assign 0 if there is no operation performed.

Write a program that would minimize the number of possible operations to transform any string x into a string y.

Input

The input consists of the strings x and y prefixed by their respective lengths, which are within 1000.

Output

An integer representing the minimum number of possible operations to transform any string x into a string y.

Sample Input

10 AGTCTGACGC
11 AGTAAGTAGGC

Sample Output

/*
题目大意：求串S变成T的最少改变次数.(变换，删除，插入)。
题目解答：Dp简单题
Source Code

Problem: 3356  User: wawadimu 
Memory: 4136K  Time: 16MS 
Language: C++  Result: Accepted 

Source Code 
*/
#include<iostream>
using namespace std;

#define maxn 1010
char a[maxn],b[maxn];
int ans[maxn][maxn];

int min(int x,int y) {return x>y ? y:x;}
int main()
{
	//freopen("3356.txt","r",stdin);
	int n,m;
	int i,j,k;
	while(scanf("%d ",&n)!=EOF)
	{
		for(i=0;i<n;i++)
		{
			//cin>>a[i];
			scanf(" %c",a+i);
			//cout<<a[i];
		}
		//cout<<a[n-1]<<endl;
		scanf("/n%d ",&m);
		for(i=0;i<m;i++)
		{
			//cin>>b[i];
			scanf("%c",b+i);
		}
		for(i=0;i<=n;i++)//注意下标
			ans[i][0]=i;
		for(j=0;j<=m;j++)
			ans[0][j]=j;
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=m;j++)
			{
				if(a[i-1]==b[j-1])
					ans[i][j]=min(min(ans[i-1][j-1],ans[i-1][j]+1),ans[i][j-1]+1);
				else
					ans[i][j]=min(min(ans[i-1][j-1]+1,ans[i-1][j]+1),ans[i][j-1]+1);
			}
		}
		printf("%d/n",ans[n][m]);
	}
	return 0;
}